$"Moles of calcium"$ $=$ $(1200xx10^-3*g)/(40.08*g*mol^-1)$ $=$ $0.04008*mol$. And thus we need $0.04008*mol$ $"calcium carbonate"$, i.e. a mass of $0.04008*molxx100.09*g*mol^-1$ $=$ $??g$.
1 Answers 1 viewsSodium chloride is composed of $Na^+$ ions and $Cl^-$ ions. Now of course salt is neutral; and to ensure this sodium ion and chloride ions combine 1:1. On the other...
1 Answers 1 viewsThe balanced chemical equation The molar mass of $"CaCO"_3$ The molar ratio of $"CaO"$ to $"CaCO"_3$ The molar mass of $"CaO"$ Your strategy has four steps:...
1 Answers 1 viewsWe need a stoichiometric reaction that represents the decomposition of calcium carbonate: $CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr$ And thus calcium oxide and calcium carbonate are present in equimolar...
1 Answers 1 viewsThe idea here is that you need to use the given mass ratio to determine how much fluorine will be produced if the also produces $"34.5 g"$ of sodium....
1 Answers 1 views$MgCO_3(s)+DeltararrMgO(s) + CO_2(g)$ And we would have to heat the magnesium salt VERY fiercely to get complete decarboxylation. $"Moles of carbon dioxide LOST"$ $=$ $(4.40*g)/(44.01*g*mol^-1)$ $=0.010*mol$ And therefore in...
1 Answers 1 viewsWe interrogate the stoichiometric equation...... $CaCO_3(s) + Delta rarr CaO(s) + CO_2(g) uarr$ We require a molar quantity of $(15.0*g)/(56.08*g*mol^-1)=0.268*mol,$ with respect to $"calcium oxide"$. And clearly, we need a...
1 Answers 1 viewsThe balanced equation for this reaction: $HCl (aq) + NaF (aq) rarr HF (aq) + NaCl (aq)$ We can see that the number of moles of $HCl$ is equal to...
1 Answers 1 viewsThe first step for any chemical reaction is to write out the chemical equation to determine the molar ratios of the reactants and products. Then we will use the molecular...
1 Answers 1 viewsGiven: The process of dissolving $CaCl_2$ in releases $(8.3xx10^4"J")/("mole of "CaCl_2)$ Compute the heat energy released for $0.02"mole of "CaCl_2$: $(8.3xx10^4"J")/(cancel("mole of "CaCl_2))(0.02cancel("mole of "CaCl_2))/1 = 1660"J"$ This heat energy...
1 Answers 1 views